What am i supose to do here?
A chemist performs the following experiment:
2FeCl3 + 3CaCO3 ---> Fe2(CO3)3 + 3CaCl
If he starts with 105g FeCl3 and excess CaCO3 how many grams of CaCl2 will he obtain?
Thanks,
~Nick
how many moles of Iron(III)chloride is 105g? You get 3/2 that number of moles of CalciumChloride. Then convert that to grams.
So is the answer 74.2g? Or did i do something wrong?
To determine the number of grams of CaCl2 obtained in this chemical reaction, you need to calculate the stoichiometry of the reaction and use it to convert the given mass of FeCl3 to the desired mass of CaCl2.
1. Balance the equation:
2FeCl3 + 3CaCO3 → Fe2(CO3)3 + 3CaCl2
The equation is already balanced, but it should be CaCl2 instead of CaCl. This means that for every 2 moles of FeCl3 used, 3 moles of CaCl2 will be produced.
2. Calculate the molar mass:
Molar mass of FeCl3 = 55.8 g/mol + 3(35.45 g/mol) = 162.3 g/mol
Molar mass of CaCl2 = 40.1 g/mol + 2(35.45 g/mol) = 110.98 g/mol
3. Convert the given mass to moles:
Moles of FeCl3 = mass of FeCl3 / molar mass of FeCl3 = 105 g / 162.3 g/mol = 0.6471 mol
4. Apply stoichiometry to find the moles of CaCl2 formed:
According to the balanced equation, the mole ratio between FeCl3 and CaCl2 is 2:3.
Moles of CaCl2 = Moles of FeCl3 * (3 moles of CaCl2 / 2 moles of FeCl3) = 0.6471 mol * (3/2) = 0.9707 mol
5. Convert moles back to grams:
Mass of CaCl2 = Moles of CaCl2 * molar mass of CaCl2 = 0.9707 mol * 110.98 g/mol = 107.9 g
Therefore, the chemist will obtain approximately 107.9 grams of CaCl2.